Analogues of the elementary symmetric polynomials for the alternating group

Question

In the case of three variables, the elementary symmetric polynomials are $$ \begin{align} e_1(X_1,X_2,X_3)&:=X_1+X_2+X_3, \\ e_2(X_1,X_2,X_3)&:=X_1 X_2+X_1 X_3+X_2 X_3, \\ e_3(X_1,X_2,X_3)&:=X_1 X_2 X_3. \\ \end{align}$$ Knowledge of the values of $e_1,e_2,e_3$ determines the variables $X_1,X_2,X_3$ up to any permutation of $S_3$. That is, if $$ \begin{align} e_1(X_1,X_2,X_3)&=e_1(Y_1,Y_2,Y_3), \\ e_2(X_1,X_2,X_3)&=e_2(Y_1,Y_2,Y_3) ,\\ e_3(X_1,X_2,X_3)&=e_2(Y_1,Y_2,Y_3),\\ \end{align}$$ then there exists a permutation $\sigma \in S_3$ such that $$X_i=Y_\sigma(i) ,$$ for all $1 \leq i \leq 3$.

I'm curious as to whether there exist other "less symmetric" polynomials, say $\{P_n(X_1,X_2,X_3)\}_n$ such that having $$P_n(X_1,X_2,X_3)=P_n(Y_1,Y_2,Y_3) $$for all $n$ implies that there exists an even permutation $\sigma \in A_3 \subsetneq S_3$ for which $X_i=Y_\sigma(i)$ for all $1 \leq i \leq 3$.

I have tried keeping two of the elementary symmetric polynomials, replacing the third, but that didn't work out.

I would appreciate help with finding such polynomials $P_n$ (if they exist). Thank you!

Answer 1

Here’s one way to do it: to get a list of polynomials for any $n$, start with the list of elementary symmetric polynomials on $n$ variables and add in the Vandermonde polynomial on $n$ variables.

Answer 2

You're almost there. It suffices to add a fourth polynomial :

$$ \begin{array}{lcl} P_1 &=& X_1+X_2+X_3 \\ P_2 &=& X_1X_2+X_1X_3+X_2X_3 \\ P_3 &=& X_1X_2X_3 \\ P_4 &=& X_1^2X_2+X_2^2X_3+X_3^2X_1 \end{array} $$

Suppose that $X=(X_1,X_2,X_3)$ and $Y=(Y_1,Y_2,Y_3)$ satisfy $P_n(X)=P_n(Y)$ for $1\leq n \leq 4$. Using the first three equalities only, we already know that there is a $\sigma\in {\mathfrak S}_n$ such that $X=Y_\sigma$ (by which I mean that $X_i=Y_{\sigma(i)}$ for $1\leq i \leq 3$).

If $\sigma$ is already even, we are done. Otherwise, $\sigma$ is a transposition, say $\sigma=(1,2)$. Then the last identity becomes $P_4(Y_2,Y_1,Y_3)=P_4(Y_1,Y_2,Y_3)$. Now the polynomial $D=P_4(Y_2,Y_1,Y_3)-P_4(Y_1,Y_2,Y_3)$ factorizes as

$$ D=(Y_2-Y_1)(Y_3-Y_1)(Y_3-Y_2) $$

So at least two $Y_i$'s are equal. It is easy to deduce from here that there are other permutations $\gamma$ satisfying $X=Y_\gamma$, some of which are even.